Solve x^2/x-5+7x-8/x-5=0

$$\frac { x ^ { 2 } } { x - 5 } + \frac { 7 x - 8 } { x - 5 } = 0$$

Solve for x

$$x=-8$$

$$x=1$$

Steps Using Factoring

Variable x cannot be equal to 5 since division by zero is not defined. Multiply both sides of the equation by x-5.

$$x^{2}+7x-8=0$$

To solve the equation, factor x^{2}+7x-8 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=7 ab=-8$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -8.

$$-1,8 -2,4$$

Calculate the sum for each pair.

$$-1+8=7 -2+4=2$$

The solution is the pair that gives sum 7.

$$a=-1 b=8$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-1\right)\left(x+8\right)$$

To find equation solutions, solve x-1=0 and x+8=0.

$$x=1 x=-8$$

Steps Using Factoring By Grouping

Variable x cannot be equal to 5 since division by zero is not defined. Multiply both sides of the equation by x-5.

$$x^{2}+7x-8=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-8. To find a and b, set up a system to be solved.

$$a+b=7 ab=1\left(-8\right)=-8$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -8.

$$-1,8 -2,4$$

Calculate the sum for each pair.

$$-1+8=7 -2+4=2$$

The solution is the pair that gives sum 7.

$$a=-1 b=8$$

Rewrite x^{2}+7x-8 as \left(x^{2}-x\right)+\left(8x-8\right).

$$\left(x^{2}-x\right)+\left(8x-8\right)$$

Factor out x in the first and 8 in the second group.

$$x\left(x-1\right)+8\left(x-1\right)$$

Factor out common term x-1 by using distributive property.

$$\left(x-1\right)\left(x+8\right)$$

To find equation solutions, solve x-1=0 and x+8=0.

$$x=1 x=-8$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 5 since division by zero is not defined. Multiply both sides of the equation by x-5.

$$x^{2}+7x-8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 7 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-7±\sqrt{7^{2}-4\left(-8\right)}}{2}$$

Square 7.

$$x=\frac{-7±\sqrt{49-4\left(-8\right)}}{2}$$

Multiply -4 times -8.

$$x=\frac{-7±\sqrt{49+32}}{2}$$

Add 49 to 32.

$$x=\frac{-7±\sqrt{81}}{2}$$

Take the square root of 81.

$$x=\frac{-7±9}{2}$$

Now solve the equation x=\frac{-7±9}{2} when ± is plus. Add -7 to 9.

$$x=\frac{2}{2}$$

Divide 2 by 2.

$$x=1$$

Now solve the equation x=\frac{-7±9}{2} when ± is minus. Subtract 9 from -7.

$$x=-\frac{16}{2}$$

Divide -16 by 2.

$$x=-8$$

The equation is now solved.

$$x=1 x=-8$$

Steps for Completing the Square

Variable x cannot be equal to 5 since division by zero is not defined. Multiply both sides of the equation by x-5.

$$x^{2}+7x-8=0$$

Add 8 to both sides. Anything plus zero gives itself.

$$x^{2}+7x=8$$

Divide 7, the coefficient of the x term, by 2 to get \frac{7}{2}. Then add the square of \frac{7}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+7x+\left(\frac{7}{2}\right)^{2}=8+\left(\frac{7}{2}\right)^{2}$$

Square \frac{7}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+7x+\frac{49}{4}=8+\frac{49}{4}$$

Add 8 to \frac{49}{4}.

$$x^{2}+7x+\frac{49}{4}=\frac{81}{4}$$

Factor x^{2}+7x+\frac{49}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{7}{2}\right)^{2}=\frac{81}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{2}\right)^{2}}=\sqrt{\frac{81}{4}}$$

Simplify.

$$x+\frac{7}{2}=\frac{9}{2} x+\frac{7}{2}=-\frac{9}{2}$$

Subtract \frac{7}{2} from both sides of the equation.

$$x=1 x=-8$$